Question

Range of sin‐¹x+cos-¹x+ tan‐¹ x is​

Answer 1

Answer:

The given function is…

f(x)=sin−1x+cos−1x+tan−1xf(x)=sin−1x+cos−1x+tan−1x

We know that…

sin−1x+cos−1x=π2sin−1x+cos−1x=π2

So,f(x)=sin−1x+cos−1x+tan−1x=π2+tan−1xf(x)=sin−1x+cos−1x+tan−1x=π2+tan−1x

Since,the range of tan−1xtan−1x is [−π2,π2][−π2,π2]

So we can write….

−π2≤tan−1x≤π2⇒π2−π2≤π2

Step-by-step explanation:

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